Not Homework. Proving question

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The discussion centers on proving the limit of the expression \((\frac{10^{n}}{10^{n}-1})^{10^n} = e\) as \(n\) approaches infinity. Participants suggest simplifying the expression by substituting \(u = 10^n\) and analyzing the limit \(\lim_{u \to \infty}\left( \frac{u}{u - 1} \right)^u\). The approach involves taking the natural logarithm of both sides, leading to the equation \(ln y = u ln(u/(u - 1))\), and subsequently evaluating the limit of this logarithmic expression.

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Homework Statement



Prove that:

[tex](\frac{10^{n}}{10^{n}-1})^{10^n} = e[/tex]

as n approaches inf.

Homework Equations



It is rather obvious that i can let 10^{n} be yet another variable


The Attempt at a Solution



I proved by assuming binomial dist as Poisson dist (it is interesting for me to use this to prove a continued fraction)
 
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icystrike said:

Homework Statement



Prove that:

[tex](\frac{10^{n}}{10^{n}-1})^{10^n} = e[/tex]

as n approaches inf.

Homework Equations



It is rather obvious that i can let 10^{n} be yet another variable


The Attempt at a Solution



I proved by assuming binomial dist as Poisson dist (it is interesting for me to use this to prove a continued fraction)

You can simplify by letting u = 10n and working with this limit:

[tex]\lim_{u \to \infty}\left( \frac{u}{u - 1} \right)^u[/tex]

The usual approach is to let y = (u/(u - 1))u, and then take the natural log of both sides.

ln y = u ln(u/(u - 1)) = [ln(u/(u - 1))]/(1/u)

Now take the limit of both sides. Note that the expression above is the limit of the log of what you want.
 

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